Theorem ndmovg 7595

Description: The value of an operation outside its domain. (Contributed by NM, 28-Mar-2008.)

Assertion

Ref	Expression
ndmovg	⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Proof of Theorem ndmovg

Step	Hyp	Ref	Expression
1		df-ov 7415	. 2 ⊢ (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2		eleq2 2851	. . . . . 6 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆)))
3		opelxp 5696	. . . . . 6 ⊢ (⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆) ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆))
4	2, 3	bitrdi 290	. . . . 5 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
5	4	notbid 321	. . . 4 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
6		ndmfv 6913	. . . 4 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
7	5, 6	biimtrrdi 257	. . 3 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅))
8	7	imp 411	. 2 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
9	1, 8	eqtrid 2809	1 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 400   = wceq 1569   ∈ wcel 2142  ∅c0 4285  ⟨cop 4594   × cxp 5658  dom cdm 5660  ‘cfv 6536  (class class class)co 7412

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5256  ax-nul 5268  ax-pr 5403

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-opab 5173  df-xp 5666  df-dm 5670  df-iota 6492  df-fv 6544  df-ov 7415

This theorem is used by:  ndmov  7596  curry1val  8098  curry2val  8102  1div0  11879  repsundef  14815  cshnz  14836  mamufacex  22564  mavmulsolcl  22719  mavmul0g  22721  iscau2  25447  1div0apr  30830  rrxsphere  49556